At the Introductory Physics page where you posted this, there is a sticky at the top with different formulas in there. The link is <https://www.physicsforums.com/showthread.php?t=110015> and the formula you need is in there. It's in the second post that Doc Al posted under "General".
You have...
It's been awhile since I've done any of this so somebody correct me if otherwise, but for west of north and for above the horizontal, you should not be using any distance as the hypotenuse.
You will need two different triangles for each situation which you did, but the hypotenuse is not...
I had this problem or almost exact one back in high school. I found it easy to draw a picture, convert each vector into vertical and horizontal components...
I think I get it, so you just calculated the hypotenuse of the triangle to be 60 as well? Which you didn't really need to do, but it's good to double check too. But yes, the tree I got was 30.02m >> 30m.
You should get an answer of 30.02m which makes sense. If the shadow was the same as the height of the tree, that would suggest the angle is 45 (middle of 0-90). If the shadow was shorter than the tree, it would suggest the sun be more overtop of the tree, giving an angle closer to 90...
You were right with the 52, I managed to put a crappy diagram there, but O is the sun, and that line ' - ' is the line that makes an angle of 30 degrees with the sun and the horizontal. So you say the angle should be in the corner A, but does the diagram help you to determine if you are right...